How many solutions of the equation $\tan x = \tan (\tan x)$ are on the interval $0 \le x \le \tan^{-1} 942$?  (Here $\tan^{-1}$ means the inverse tangent function, sometimes written $\arctan$.)

Note: You can assume the result that $\tan \theta > \theta$ for $0 < \theta < \frac{\pi}{2}.$
Solution: Two angles have the same tangent if and only if they differ by a multiple of $\pi.$  This means $\tan x - x$ is a multiple of $\pi.$  Let
\[T(x) = \tan x - x.\]First, we prove that the function $T(x)$ is strictly increasing on the interval $\left[ 0, \frac{\pi}{2} \right).$  Let $0 \le x < y < \frac{\pi}{2}.$  Then
\[y - x < \tan (y - x) = \frac{\tan y - \tan x}{1 + \tan x \tan y} \le \tan y - \tan x.\]Re-arranging, we get $\tan x - x < \tan y - y,$ or $T(x) < T(y).$

Note that as $x$ approaches $\frac{\pi}{2},$ $T(x)$ approaches infinity.  This means for every nonnegative integer $n,$ there exists a unique value of $x$ such that $T(x) = n \pi.$

We have the estimate $300 \pi \approx 942.48.$  Hence,
\[T(\tan^{-1} 942) = 942 - \tan^{-1} 942 < 942 < 300 \pi.\]Also,
\[T(\tan^{-1} 924) = 942 - \tan^{-1} 942 > 942 - \frac{\pi}{2} > 299 \pi.\]Since $299 \pi < T(\tan^{-1} 942) < 300 \pi,$ the equation $T(x) = n \pi$ has a solution on the interval $[0, \tan^{-1} 942]$ if and only if $0 \le n < 300,$ so there are $\boxed{300}$ solutions.